80240233: Computational Complexity Lecture 3 ITCS, Tsinghua Univesity, Fall 2007 16 October 2007 Instructor: Andrej Bogdanov Notes by: Jialin Zhang and Pinyan Lu In this lecture, we introduce the complexity class coNP, the Polynomial Hierarchy and the notion of oracle. 1 The class coNP The complexity class NP contains decision problems asking this kind of question: On input x, ∃y : |y| = p(|x|) s.t.V (x, y), where p is a polynomial and V is a relation which can be computed by a polynomial time Turing Machine (TM). Some examples: SAT: given boolean formula φ, does φ have a satisfiable assignment? MATCHING: given a graph G, does it have a perfect matching? Now, consider the opposite problems of SAT and MATCHING. UNSAT: given boolean formula φ, does φ have no satisfiable assignment? UNMATCHING: given a graph G, does it have no perfect matching? This kind of problems are called coNP problems. Formally we have the following definition. Definition 1. For every L ⊆ {0, 1}∗, we say that L ∈ coNP if and only if L ∈ NP, i.e. L ∈ coNP iff there exist a polynomial-time TM A and a polynomial p, such that x ∈ L ⇔ ∀y, |y| = p(|x|) A(x, y) accepts. Then the natural question is how P, NP and coNP relate. First we have the following trivial relationships. Theorem 2. P ⊆ NP, P ⊆ coNP For instance, MATCHING ∈ coNP. And we already know that SAT ∈ NP. Is SAT ∈ coNP? If it is true, then we will have NP = coNP. This is the following theorem. Theorem 3. If SAT ∈ coNP, then NP = coNP. 1 2 Proof. Take any L ∈ NP, we can reduce L to SAT. Since SAT ∈ coNP, so there exists an coNP algorithm for SAT. Therefore, there exists an coNP algorithm for L. So L ∈ coNP. So we proved that NP ⊆ coNP. For the other side, we have L ∈ coNP ⇒ L ∈ NP ⇒ L ∈ coNP ⇒ L ∈ NP. The first and last steps use the definition of coNP and the middle step use the result NP ⊆ coNP, which is proved above. To sum up, we complete the proof. 2 Polynomial Hierarchy Here we consider a new problem MIN-EQUIV. Given a boolean formula φ, is φ the smallest formula that computes the function φ? Formally, φ ∈ MIN-EQUIV ⇔ ∀φ0 < φ, ∃x : φ0(x) 6= φ(x). The opposite problem of MIN-EQUIV is MIN-EQUIV. Given a boolean formula φ, is there a smaller formula that compute the function φ? There is no obvious notion of a certificate of membership. It seems that the way to capture such languages is to allow not only an ”exists” quantifier (as in the definition of NP) or only a ”for all” quantifier (as in the definition of coNP). This motivates the following definition: Definition 4. Σ2 is defined to be the class of decision problems for which there exists a polynomial- time TM A and a polynomial p such that x ∈ L ⇔ ∃y1∀y2 A(x, y1, y2) accepts, where |y1| = p(|x|), |y2| = p(|x|). Π2 is defined to be the class of decision problems for which there exists a polynomial-time TM A and a polynomial p such that x ∈ L ⇔ ∀y1∃y2 A(x, y1, y2) accepts, where |y1| = p(|x|), |y2| = p(|x|). The polynomial hierarchy generalizes the definitions of NP, coNP, Σ2,Π2. Definition 5. Σk is defined to be the class of decision problems for which there exists a Polynomial- time TM A and a polynomial p such that x ∈ L ⇔ ∃y1∀y2 · · · ∃/∀yk A(x, y1, y2, ··· , yk) accepts, where |yi| = p(|x|), i = 1, ··· k. Πk is defined to be the class of decision problems for which there exists a Polynomial-time TM A and a polynomial p such that x ∈ L ⇔ ∀y1∃y2 · · · ∃/∀yk A(x, y1, y2, ··· , yk) accepts, where |yi| = p(|x|), i = 1, ··· k. The polynomial hierarchy is the class PH = ∪iΣi. There are some basic observations about polynomial hierarchy: 3 1. P = Σ0 = Π0; 2. NP = Σ1, coNP = Π1; 3. P ⊆ NP ∩ coNP ⊆ Σ2, Π2 ⊆ Σ3, Π3 ⊆ · · · ; 4. L ∈ NP ⇔ L ∈ coNP; 5. L ∈ Σk ⇔ L ∈ Πk. 6. ∀k, Σk, Πk ⊆ EXP . We believe – but don’t know how to prove – that Σk 6= Σk+1,Πk 6= Πk+1,Σk 6= Πk and EXP 6= Σk, Πk for all k. There is a survey by Schaeffer and Umans[1, 2], which gives several nature complete problems for Σ2, Σ3, Π2, Π3. 3 Oracle Oracle is equivalent of subroutine in complexity theory. An oracle Turing Machine can be executed with access to a special tape, where they can make queries of the form ”is q ∈ L” for some language L and get the answer in one step. That is, oracle gives us functionality what we do not know how to implement efficiently. Definition 6. P A is defined to be all decision problems decided by Polynomial-time oracle TM given access to oracle A. Definition 7. NPA is defined bo be all decision problems decided by Nondeterministic Polynomial- time oracle TM given access to oracle A. We already saw a special kind of oracle computation, namely a reduction. In a reduction the oracle is asked only one question, and the answer to this question is the output of the algorithm. In particular, if decision problem A reduces to B, then A ∈ PB. Let’s play with oracles for a bit to get a feel about what they do: 1. What is PMATCHING? This is just P, since any call to the oracle can be simulated by the polynomial-time machine making the call. For the same reason NPMATCHING = NP. 2. How about PSAT? A polynomial-time machine with a SAT oracle can solve any NP question by Cook’s theorem – first reduce to SAT then ask the question. But it can also solve any coNP question – again, reduce to SAT, ask the question, then output the opposite answer. So we have PSAT ⊇ NP, coNP. SAT 3. Since P is more powerful than both NP and coNP, how does it relate to Σ2 and Π2? The SAT following theorem, implies that P ⊆ Σ2 ∩ Π2. 4 SAT Theorem 8. NP = Σ2 SAT Proof. Step 1: prove Σ2 ⊆ NP . For any L ∈ Σ2, there exists a polynomial time TM V such that x ∈ L ⇔ ∃y∀z V (x, y, z) accepts. We can think ”∃y” part to be the nondeterministic tape of a NTM N. Once N guesses y, it has to determine whether ∀z : V (x, y, z) accepts. We can ask SAToracle here, ”does there exists z, s.t.V (x, y, z) rejects?” and output the opposite answer. SAT SAT Step 2: prove NP ⊆ Σ2. For any L ∈ NP , we are given an oracle Polynomial-time NTM N. We need to simulate N SAT by ∃y, ∀z, V (x, y, z) accepts. Nondeterministic tape of N is part of y in ”∃y” of V . When N makes an oracle call Φi, V keeps trace of Φi, and guesses an answer ai to Φi . In the end, V will check ”Yes” answers(φi ∈ SAT) and ”No” answers (φi ∈ UNSAT). To sum up, we can define the Σ2 language as following: ∃y∃a1, a2, ··· , ak∃v1, v2, ··· vk∀w1, w2, ··· , wk V (x, y, a, v, w)accepts, where V (x, y, a, v, w) accepts if and only if the following two conditions satisfy: (1) given input x, nondeterministic tape y and the oracle’s answer a, N(x, y, a) accepts; (2) ai is a correct answer of Φi, which means that for every i, 1 ≤ i ≤ k, either ai = ”Yes” and Φi(vi) = ”Yes” or ai = ”No” and Φi(wi) = ”No”. This argument gives an alternate characterization of the polynomial hierarchy using oracle ma- chines, and it can be extended to higher levels of the hierarchy too: SAT 1. coNP = Π2; Σk−complete problem 2. Σk+1 = NP . ΣkSAT Since ”ΣkSAT” is complete for class Σk, this means NP = Σk+1. The following table summarizes some conjectures people believe and the fact people prove about polynomial hierarchy. P, NP, coNP PH Believe P 6= NP Σk 6= Σk+1 for all k NP 6= coNP Σk 6= Πk for all k Fact P = NP ⇔ P = coNP Σk = Σk+1 ⇒ Πk+1 = Σk+1 = Πk = Σk Theorem 9. P = NP ⇒ Σ2 = P 5 Proof. Take any L ∈ Σ2. This means x ∈ L ⇔ ∃y∀z, V (x, y, z) Accepts. We define another language L0 :(x, y) ∈ L0 ⇒ ∀z0,V (x, y, z0) accepts. So L0 ∈ coNP. By the assumption P = NP, we have L0 ∈ P. This means there exists Polynomial-time TM V such that (x, y) ∈ L0 ⇔ V 0(x, y) accepts. Then, x ∈ L ⇒ ∃y V 0(x, y) accepts. This means L ∈ NP = P. References [1] M. Schaeffer and C. Umans. Completeness in the Polynomial-Time Hierarchy: a compendium. SIGACT News. guest Complexity Theory column. September 2002. [2] M. Schaeffer and C. Umans. Completeness in the Polynomial-Time Hierarchy: Part II. SIGACT News. guest Complexity Theory column. December 2002..